Answer:
v = 7.69 x 10³ m/s = 7690 m/s
T = 5500 s = 91.67 min = 1.53 h
Explanation:
In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:
![F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\](https://tex.z-dn.net/?f=F_%7Bgravitation%7D%3D%20F_%7Bcentripetal%7D%5C%5C%5C%5C%5Cfrac%7BGM_%7Bs%7D%20M_%7BE%7D%7D%7Br%5E2%7D%20%20%3D%20%5Cfrac%7BM_%7Bs%7D%20v%5E2%7D%7Br%7D%5C%5C%5C%5C%5Cfrac%7BGM_%7BE%7D%7D%7Br%7D%20%3D%20v%5E2%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7BGM_%7BE%7D%7D%7Br%7D%20%7D%20%5C%5C%5C%5C)
where,
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Me = Mass of Earth = 5.97 x 10²⁴ kg
r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m
v = orbital speed = ?
Therefore,
![v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B%286.67%20x%2010%5E%7B-11%7DN.m%5E2%2Fkg%5E2%29%285.97%20x%2010%5E%7B24%7D%20kg%29%7D%7B6.732%20x%2010%5E6%20m%7D%20%7D%5C%5C%5C%5C)
<u>v = 7.69 x 10³ m/s</u>
For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.
So, its orbital speed can be given as:
![v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7BCircumference%20of%20Circle%20at%20Given%20Altitude%7D%7BT%7D%5C%5C%5C%5Cv%20%3D%20%20%5Cfrac%7B2%5Cpi%20r%7D%7BT%7D%5C%5C%5C%5C)
where,
T = Time Period of Satellite = ?
Therefore,
![T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20r%7D%7Bv%7D%5C%5C%5C%5CT%20%3D%20%5Cfrac%7B%282%5Cpi%20%29%286.732%20x%2010%5E6%20m%7D%7B7.69%20x%2010%5E3%20m%2Fs%7D%5C%5C%5C%5C)
<u>T = 5500 s = 91.67 min = 1.53 h</u>
Answer: magnitude of the instantaneous angular velocity
Explanation:
Instantaneous angular speed is refered to as the magnitude of the instantaneous angular velocity. We should note that the instantaneous angular velocity is the rate that has to do with the rotation of an object in circular path.
In physics the standard unit of weight is Newton, and the standard unit of mass is the kilogram. On Earth, a 1 kg object weighs 9.8 N, so to find the weight of an object in N simply multiply the mass by 9.8 N. Or, to find the mass in kg, divide the weight by 9.8 N.
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N